Geometry find length given altitude - Is my understanding correct?
$begingroup$
Question: For the the triangle $ABC$ , $BC=10$ and $AB=AC=13$. Given that $AD$ is a median and $BE$ is an altitude, find what $DE$ is equal to.
Well I construct the triable and I know that AD breaks BC into two equal parts of $5$. And that both $BE$ and $AD$ form right triangles.
If I look at triangle $BCE$, with angle $E$ being $90$ degrees, how can I say that $DE=DC=5=BD$
That's the answer, but I'm unsure as to why. I was leaning to thales theorem, where I can see $BDC$ as a diameter of a circle, that has angle $E$ being $90$ degrees. So they're all radius on the circle? Thus equal to 5?
geometry euclidean-geometry
edited Dec 5 '18 at 5:36
random123
1,2601720
asked Dec 5 '18 at 3:26
OnePieceFan123OnePieceFan123
62
$endgroup$
add a comment |
$begingroup$
Question: For the the triangle $ABC$ , $BC=10$ and $AB=AC=13$. Given that $AD$ is a median and $BE$ is an altitude, find what $DE$ is equal to.
Well I construct the triable and I know that AD breaks BC into two equal parts of $5$. And that both $BE$ and $AD$ form right triangles.
If I look at triangle $BCE$, with angle $E$ being $90$ degrees, how can I say that $DE=DC=5=BD$
That's the answer, but I'm unsure as to why. I was leaning to thales theorem, where I can see $BDC$ as a diameter of a circle, that has angle $E$ being $90$ degrees. So they're all radius on the circle? Thus equal to 5?
geometry euclidean-geometry
edited Dec 5 '18 at 5:36
random123
1,2601720
asked Dec 5 '18 at 3:26
OnePieceFan123OnePieceFan123
62
$endgroup$
$begingroup$
You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
1
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41
add a comment |
$begingroup$
Question: For the the triangle $ABC$ , $BC=10$ and $AB=AC=13$. Given that $AD$ is a median and $BE$ is an altitude, find what $DE$ is equal to.
Well I construct the triable and I know that AD breaks BC into two equal parts of $5$. And that both $BE$ and $AD$ form right triangles.
If I look at triangle $BCE$, with angle $E$ being $90$ degrees, how can I say that $DE=DC=5=BD$
That's the answer, but I'm unsure as to why. I was leaning to thales theorem, where I can see $BDC$ as a diameter of a circle, that has angle $E$ being $90$ degrees. So they're all radius on the circle? Thus equal to 5?
geometry euclidean-geometry
edited Dec 5 '18 at 5:36
random123
1,2601720
asked Dec 5 '18 at 3:26
OnePieceFan123OnePieceFan123
62
$endgroup$
Question: For the the triangle $ABC$ , $BC=10$ and $AB=AC=13$. Given that $AD$ is a median and $BE$ is an altitude, find what $DE$ is equal to.
Well I construct the triable and I know that AD breaks BC into two equal parts of $5$. And that both $BE$ and $AD$ form right triangles.
If I look at triangle $BCE$, with angle $E$ being $90$ degrees, how can I say that $DE=DC=5=BD$
That's the answer, but I'm unsure as to why. I was leaning to thales theorem, where I can see $BDC$ as a diameter of a circle, that has angle $E$ being $90$ degrees. So they're all radius on the circle? Thus equal to 5?
geometry euclidean-geometry
geometry euclidean-geometry
edited Dec 5 '18 at 5:36
random123
1,2601720
asked Dec 5 '18 at 3:26
OnePieceFan123OnePieceFan123
62
edited Dec 5 '18 at 5:36
random123
1,2601720
asked Dec 5 '18 at 3:26
OnePieceFan123OnePieceFan123
62
edited Dec 5 '18 at 5:36
random123
1,2601720
edited Dec 5 '18 at 5:36
random123
1,2601720
edited Dec 5 '18 at 5:36
random123
1,2601720
1,2601720
asked Dec 5 '18 at 3:26
OnePieceFan123OnePieceFan123
62
asked Dec 5 '18 at 3:26
OnePieceFan123OnePieceFan123
62
asked Dec 5 '18 at 3:26
OnePieceFan123OnePieceFan123
62
62
$begingroup$
You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
1
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41
add a comment |
$begingroup$
You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
1
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41
$begingroup$
You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
1
1
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026563%2fgeometry-find-length-given-altitude-is-my-understanding-correct%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026563%2fgeometry-find-length-given-altitude-is-my-understanding-correct%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
1
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41